Question: $f(t) = -t^{2}-6t+g(t)$ $h(t) = -4t^{2}-3t-3(g(t))$ $g(n) = -2n-7$ $ f(g(-9)) = {?} $
Explanation: First, let's solve for the value of the inner function, $g(-9)$ . Then we'll know what to plug into the outer function. $g(-9) = (-2)(-9)-7$ $g(-9) = 11$ Now we know that $g(-9) = 11$ . Let's solve for $f(g(-9))$ , which is $f(11)$ $f(11) = -11^{2}+(-6)(11)+g(11)$ To solve for the value of $f$ , we need to solve for the value of $g(11)$ $g(11) = (-2)(11)-7$ $g(11) = -29$ That means $f(11) = -11^{2}+(-6)(11)-29$ $f(11) = -216$